Find the min of z

The given expression can be factored as $z = (x + y + 1)^{2} + 2(y + 1)^{2} + 1$ , which has a minimum of $\boxed{1}$ when both $x + y + 1 = 0$ and $y + 1 = 0$ , i.e., when $y = -1, x = 0$ .

If $z = f(x,y)$ , then we can calculate $grad(f) = 0$ . This yields the partial derivatives:

$f_{x} = 2x + 2y + 2 = 0;$

$f_{y} = 2x + 6y + 6 = 0$

or $(x,y) = (0,-1)$ as the critical point. The Hessian matrix $F(x,y)$ at this critical point computes to:

$F(0,-1) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 6 \end{bmatrix}$

which is a positive-definite matrix for ALL points $(x,y)$ and yields a global minimum for $(0,-1)$ . Thus the minimum value is just $z = f(0,-1) = \boxed{1}.$